three consecutive integers are such that when they are taken in increasing order and multiplied by 2,3 and 4 respectivly, they add up to 74 . find these numbers
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Answer:
Three consecutive integers are such that when they are taken in increasing order and multiplied by 2, 3, and 4 respectively, they add up to 74. Find these numbers.
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Hint: Here, we have to find the three consecutive integers. We will assume the smallest of the three consecutive integers to be \[x\]. Using the given information, we will form a linear equation in terms of \[x\]. We will solve the obtained equation to find the value of \[x\], and hence, the three consecutive integers.
Complete step-by-step answer:
Let the smallest integer of the three consecutive integers be \[x\].
Therefore, the next two consecutive integers will be \[x + 1\] and \[x + 2\].
First, we will arrange these in increasing order.
Therefore, we get \[x\], \[x + 1\], and \[x + 2\].
Now, we will use the given information to form a linear equation in terms of \[x\].
The three consecutive integers are multiplied by 2, 3, and 4 respectively.
Multiplying \[x\] by 2, we get \[2x\].
Multiplying \[x + 1\] by 3, we get \[3\left( {x + 1} \right)\].
Multiplying \[x + 2\] by 4, we get \[4\left( {x + 2} \right)\].
It is given that the three consecutive integers multiplied by 2, 3, and 4 respectively, add up to 74.
Therefore, we can form the equation
\[2x + 3\left( {x + 1} \right) + 4\left( {x + 2} \right) = 74\]
We will solve this equation to get the value of \[x\].
Multiplying the terms of the expression, we get
\[ \Rightarrow 2x + 3x + 3 + 4x + 8 = 74\]
Adding the like terms of the expression, we get
\[ \Rightarrow 9x + 11 = 74\]
Subtracting 11 from both sides of the equation, we get
\[\begin{array}{l} \Rightarrow 9x + 11 - 11 = 74 - 11\\ \Rightarrow 9x = 63\end{array}\]
Dividing both sides by 9, we get
\[\begin{array}{l} \Rightarrow \dfrac{{9x}}{9} = \dfrac{{63}}{9}\\ \Rightarrow x = 7\end{array}\]
Therefore, the smallest consecutive integer out of the three numbers is 7.
Substituting \[x = 7\] in \[x + 1\] and \[x + 2\], we get the other two integers as
\[x + 1 = 7 + 1 = 8\]
\[x + 2 = 7 + 2 = 9\]
Therefore, the three consecutive integers are 7, 8, and 9.
Note: We have used the distributive property of multiplication to find the products \[3\left( {x + 1} \right)\] and \[4\left( {x + 2} \right)\]. The distributive property of multiplication states that \[a\left( {b + c} \right) = a \cdot b + a \cdot c\].
We can verify our answer by multiplying 7, 8, 9 by 2, 3, 4 respectively and checking the sum.
Multiplying 7 by 2, we get 14.
Multiplying 8 by 3, we get 24.
Multiplying 9 by 4, we get 36.
The sum of 14, 24, and 36 is 74.
Hence, we have verified the answer.